Estimating a Population Variance

In many manufacturing processes, it is necessary to control the amount that the process varies. For example, an automobile part manufacturer must produce thousands of parts that can be used in the manufacturing process. It is imperative that the parts vary little or not at all. How might the manufacturer measure and, consequently, control the amount of variation in the car parts? A chi-square distribution can be used to construct a confidence interval for this variance.

The chi-square distribution with a $k$ degree of freedom is the distribution of a sum of the squares of $k$ independent standard normal random variables. It is one of the most widely used probability distributions in inferential statistics (e.g., in hypothesis testing or in construction of confidence intervals). The chi-squared distribution is a special case of the gamma distribution and is used in the common chi-squared tests for goodness of fit of an observed distribution to a theoretical one, the independence of two criteria of classification of qualitative data, and in confidence interval estimation for a population standard deviation of a normal distribution from a sample standard deviation. In fact, the chi-square distribution enters all analyses of variance problems via its role in the $F$-distribution, which is the distribution of the ratio of two independent chi-squared random variables, each divided by their respective degrees of freedom.

The chi-square distribution is a family of curves, each determined by the degrees of freedom. To form a confidence interval for the population variance, use the chi-square distribution with degrees of freedom equal to one less than the sample size:

$\text{d.f.} = n-1$

There are two critical values for each level of confidence:

1. The value of ${ X }_{ R }^{ 2 }$represents the right-tail critical value.
2. The value of ${ X }_{ L }^{ 2 }$represents the left-tail critical value.

Constructing a Confidence Interval

As example, imagine you randomly select and weigh 30 samples of an allergy medication. The sample standard deviation is 1.2 milligrams. Assuming the weights are normally distributed, construct 99% confidence intervals for the population variance and standard deviation.

The areas to the left and right of ${ X }_{ R }^{ 2 }$and left of ${ X }_{ L }^{ 2 }$ are:

Area to the right of ${ X }_{ R }^{ 2 } = \frac{1-0.99}{2} = 0.005$ 

Area to the left of ${ X }_{ L }^{ 2 } = \frac{1+0.99}{2} = 0.995$ 

Using the values $n=30$, $\text{d.f.} = 29$ and $c=0.99$, the critical values are 52.336 and 13.121, respectively. Note that these critical values are found on the chi-square critical value table, similar to the table used to find $z$-scores.

Using these critical values and $s=1.2$, the confidence interval for $s^2$ is as follows:

Right endpoint:

$\displaystyle \frac{(n-1)s^2}{\chi_L^2} = \frac{(30-1)(1.2)^2}{13.121} \approx 3.183$

Left endpoint:

$\displaystyle \frac{(n-1)s^2}{\chi_R^2} = \frac{(30-1)(1.2)^2}{52.336} \approx 0.798$

So, with 99% confidence, we can say that the population variance is between 0.798 and 3.183.